Section 7.1 Improper and Separable Integrals
Subsection 7.1.1 Improper Integrals
An function defined on a closed interval has to include values on the boundary of the interval. Assuming the function is piecewise continuous (as if the habit in this course), this means that the function cannot diverge to infinity near the edge of the interval. However, an function which is only defined on an open interval can have any kind of behaviour near the boundaries, including diverging to infinity. I would like to be able to handle this kind of integral as well
In single variable calculus, we used improper integrals to deal with functions with problematic behaviour near the boundary of an interval. Thankfully, this translates nicely to iterated integrals. If a function diverges to infinity near the edge of an interval, then at least one piece of the iterated integral (and perhaps more than one piece) will be a single-variable improper integral. We approach these as we did in single variable integration: with limits approaching the problematic endpoint(s).
Consider a function \(f: (0,1] \times (0,1] \rightarrow \RR\) that diverged to infinity when \(x \rightarrow 0\text{.}\) Here is the iterated integral.
\begin{equation*}
\int_I f(x,y) dA = \int_0^1 \int_0^1 f(x,y) dy dx
\end{equation*}
Since the problems happen at the edge of the \(x\) interval, the resulting improper integral uses a limit for the \(x\) part of the iterated integral.
\begin{equation*}
\int_I f(x,y) dA = \lim_{a \rightarrow 0^+} \int_a^1 \int_0^1
f(x,y) dy dx
\end{equation*}
If, instead, the function diverged to infinity as \(y
\rightarrow 0\text{,}\) then the limit would be in the \(y\) part of the iterated integral.
\begin{equation*}
\int_I f(x,y) dA = \int_0^1 \left[ \lim_{a \rightarrow 0^+}
\int_a^1 f(x,y) dy \right] dx
\end{equation*}
Improper integrals also allow us to consider infinite domains of integral. For example, if I want to integrate over all of \(\RR^2\text{,}\) I set it us as follows.
\begin{equation*}
\int_{\RR^2} f(x,y) dA = \lim_{a \rightarrow \infty} \lim_{b
\rightarrow \infty} \int_{[-a,a]\times[-b,b]} f(x,y) dA
\end{equation*}
Then I can write this as in iterated integral where both pieces are improper integrals.
\begin{equation*}
\int_{\RR^2} f(x,y) dA = \lim_{a \rightarrow \infty}
\int_{-a}^a \left[ \lim_{b \rightarrow \infty} \int_{-b}^b
f(x,y) dy \right] dx
\end{equation*}
Consider the function \(f(x,y) = \frac{1}{\sqrt{(x^2 +
y^2)^3}}\) on the interval \(I = (0,1] \times (0,1]\text{.}\) As I approach \((0,0)\text{,}\) on the corner of the interval, the function is diverges to infinity. I set this up as an interated integral and, when necessary, I use a limit for the single-variable improper integral.
\begin{align*}
\int_I f(x,y) dA \amp = \int_0^1 \int_0^1
\frac{1}{\sqrt{(x^2 + y^2)^3}} dx dy\\
\amp = \int_0^1 \left. \frac{x}{y^2 \sqrt{x^2 + y^2}}
\right|_0^1 dy\\
\amp = \int_0^1 \frac{1}{y^2 \sqrt{1 + y^2}} dy\\
\amp = \lim_{a \rightarrow 0^+} \int_a^1 \frac{1}{y^2
\sqrt{1 + y^2}} dy\\
\amp = \lim_{a \rightarrow 0^+} \left.
\frac{-\sqrt{1+y^2}}{y} \right|_a^1 = \frac{-\sqrt{2}}{1}
+ \lim_{a \rightarrow 0^+} \frac{\sqrt{1+a^2}}{a} =
\infty
\end{align*}
This integral diverges; the volume under the curve grows to infinity as the domain includes more and more of the interval approaching the point \((0,0)\text{.}\) Improper integrals, of course, can diverge.
Example 7.1.2.
Here is an odd application of multiple integration. I know that \(e^{-x^2}\) has no elementary antiderivative. Therefore, the intergal
\begin{equation*}
A = \int_{-\infty}^\infty e^{-x^2} dx
\end{equation*}
cannot be evaluated directly. However, this is a very important integral: \(e^{-x^2}\) is the normal distribution and, in Statistics, I need to integrate it frequently. I’ll use integrals over \(\RR^2\text{,}\) strangely enough, to calculate this integral by squaring the single variable integral.
\begin{align*}
A \amp = 2 \int_{0}^\infty e^{-x^2} dx\\
A^2 \amp = \left( 2 \int_{0}^\infty e^{-x^2} dx \right)
\left(2 \int_{0}^\infty e^{-y^2} dy \right)
\end{align*}
The second integral uses a new variables since variables of integration only matter inside the integral. Then I can combine the two integrals.
\begin{align*}
A^2 \amp = 4 \int_0^\infty \int_0^\infty e^{-x^2} e^{-y^2}
dx dy\\
\amp = 4 \int_0^\infty \int_0^\infty e^{-(x^2 + y^2)} dx
dy
\end{align*}
Now I am going to do a substitution in the \(y\) variable. Treating the \(x\) variable as a constant, I replace \(y\) with \(y = xs\) so that \(dy = x ds\text{.}\) If \(y=0\) then \(s=0\) and as \(y \rightarrow
\infty\text{,}\) \(s \rightarrow \infty\text{,}\) so the bounds for \(s\) remain the same as the bounds for \(y\text{.}\) Remember, \(x\) is a constant through this whole substitution. (Using Fubini’s Theorem, I could think of \(y\) as the inside integral for this substitution, and then switch back to thinking of \(x\) as the inside integral after the substitution.)
\begin{align*}
A^2 \amp = 4 \int_0^\infty \int_0^\infty e^{-x^2 (1+s^2)}
x dx ds\\
\amp = 4 \int_0^\infty \left. \frac{-1}{2(1+s^2)}
e^{-x^2(1+s^2)} \right|_0^\infty ds\\
\amp = 2 \int_0^\infty \frac{1}{1+s^2} ds\\
\amp = \lim_{a \rightarrow \infty} 2 \arctan a - \arctan 0
= \frac{2\pi}{2} = \pi\\
A \amp = \int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}
\end{align*}
I recover the area under the bell curve: \(\sqrt{\pi}\text{.}\) It’s a very strange result. However, if you taken any statistics and worked on normal distributions, likely you will recall the presence of these strange \(\sqrt{\pi}\) terms. Now I know they are present to normalize the area (since we want a probability function to have area one under its graph).
Subsection 7.1.2 Separable Integrals
I didn’t explicitly do an example of this type, but it is useful to mention that a particular class of integrable functions have a very nice integration property.
Definition 7.1.3.
If an integrable function on \(I = [a,b] \times
[c,d]\) has the form \(f(x,y) = g(x) h(y)\) for some single variable functions \(g\) and \(h\text{,}\) then it is called separable.
For the benefit of those who have taken differential equations, this use of the term ‘separable’ matches with the use of the term for separable differential equations; a differential equation is a separable equation if and onl if \(f(x,y)\) is a separable function in the sens od this definition.
\begin{equation*}
\frac{dy}{dx} = f(x,y)
\end{equation*}
Separable functions have nice behaviour for interated integrals. If \(f(x,y) = g(x)h(y)\) is a separable function, then its integral separates entirely into two single variable integrals.
\begin{align*}
\int_I f(x,y) dy dx \amp = \int_a^b \int_c^d g(x) h(y) dy dx\\
\amp = \int_a^b g(x) \left[ \int_c^d h(y) dy \right] dx\\
\amp = \left[ \int_c^d h(y) dy \right] \left[ \int_a^b
g(x) dx \right]
\end{align*}
The operations in this calculation are justified because the pieces which I move in and out of a integral do not involve that variable of integration. In this way, I can separate the integral into two single variable integrals.
